What is the inverse of f(x)=-x/((16-x^2)^{1/2)} ?

The inverse of f(x)=-x/((16-x^2)^{1/2)} is (4x)/((x^2+1)^{1/2)},-(4x)/((x^2+1)^{1/2)}

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inverse of f(x)=-x/((16-x^2)^{1/2)}

inverse

f (x) = - \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

\frac{4 x}{( x ^{2} + 1 ) ^{\frac{1}{2}}}, - \frac{4 x}{( x ^{2} + 1 ) ^{\frac{1}{2}}}

Solution steps

y = - \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

Replace

x

with

y

x = - \frac{y}{( 16 - y ^{2} ) ^{\frac{1}{2}}}

Solve for

y, x = - \frac{y}{( 16 - y ^{2} ) ^{\frac{1}{2}}}

\frac{4 x}{( x ^{2} + 1 ) ^{\frac{1}{2}}}, - \frac{4 x}{( x ^{2} + 1 ) ^{\frac{1}{2}}}

(ln (x))^{6}

f (x) = 15 x + 4

f (x) = 15 x + 9

f (x) = - 0.1666 + 4.333

y = 200 x - 4000

What is the inverse of f(x)=-x/((16-x^2)^{1/2)} ?
The inverse of f(x)=-x/((16-x^2)^{1/2)} is (4x)/((x^2+1)^{1/2)},-(4x)/((x^2+1)^{1/2)}